Math, asked by Arpita102028, 5 months ago

1)ABCD is a cyclic quadrilateral. If <ACD = 40°, <ADC = 80°. Find <CBD
2) If the ratio of the radius of two cylinder of same height is 3/4. Find the ratio of their volume​

Answers

Answered by Anonymous
72

1) \sf{\huge{\red{\star}}} Given:-

  • ABCD is a cyclic quadrilateral.
  • \sf{\angle ACD = 40^\circ}
  • \sf{\angle ADC = 80^\circ}

\sf{\huge{\red{\star}}} To Find:-

\sf{\angle CBD}

\sf{\huge{\red{\star}}} Note:-

Refer to the attachment provided for a better idea of the scene.

\sf{\huge{\red{\star}}} Solution:-

In quad.ABCD,

\sf{\angle ACD = 40^\circ}

\sf{\angle ADC = 80^\circ}

We know,

The opposite angles of a cyclic quadrilateral are supplementary.

Hence,

\sf{\angle ADC + \angle ABC = 180^\circ}

= \sf{80^\circ + \angle ABC = 180^\circ}

= \sf{\angle ABC = 180^\circ - 80^\circ}

= \sf{\angle ABC = 100^\circ}

Now,

We also know,

Angles on the same segment of a cyclic quadrilateral are always equal.

Hence,

\sf{\angle ABD = \angle ACD}

=> \sf{\angle ABD = 40^\circ}

Now,

\sf{\angle ABD + \angle CBD = \angle ABC}

= \sf{40^\circ + \angle CBD = 100^\circ}

= \sf{\angle CBD = 100^\circ - 40^\circ}

= \sf{\angle CBD = 60^\circ}

\sf{\bold{\therefore\:\angle CBD = 60^\circ}}

______________________________________

2. \sf{\huge{\red{\star}}} Given:-

  • Ratio of radius of two cylinders = 3:4
  • Height of both the cylinders are same.

\sf{\huge{\red{\star}}} To Find:-

Ratio of volume of the two cylinders.

\sf{\huge{\red{\star}}} Assumption:-

Let the radius of 1st cylinder be \sf{r_1} and the radius of 2nd cylinder be \sf{r_2}

Let the height of both the cylinders be h.

\sf{\huge{\red{\star}}} Solution:-

We know,

Volume of a cylinder = πr²h

Therefore for 1st cylinder,

\sf{Volume\:(V_1) = \pi (r_1)^2 h}

= \sf{V_1 = \pi (3)^2 h}

= \sf{V_1 = \pi 9 h}

Now,

For the 2nd cylinder,

\sf{Volume\:(V_2) = \pi (r_2)^2 h}

= \sf{V_2 = \pi (4)^2 h}

= \sf{V_2 = \pi 16 h}

Now,

Ratio of the Volumes of both the cylinder =

\sf{Volume\:of\:1st\:cylinder\:(V_1):Volume\:of\:2nd\:cylinder\:(V_2)}

= \sf{\pi 9h : \pi 16h}

= \sf{\dfrac{\pi 9 h}{\pi 16h}}

= \sf{\dfrac{\cancel{\pi} 9 \cancel{h}}{\cancel{\pi} 16 \cancel{h}}}

= \sf{\dfrac{9}{16}}

= \sf{9:16}

Therefore the ratio of volumes of both the cylinders is 9:16.

______________________________________

Attachments:
Answered by Anonymous
23

\large{\underline{\purple{\frak{ \: Given:- \: }}}} \\

  • \sf \angle ACD = 40^{\circ}\\
  • \sf \angle ADC = 80^{\circ}\\

\large{\underline{\purple{\frak{ \: Find:- \: }}}} \\

  • \sf Value\:of\:\angle CBD\\

\large{\underline{\purple{\frak{ \: Diagram:- \: }}}} \\

\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3) \qbezier(1.7,1.6)(1.7,2)(1.7,-1.6)\qbezier(-1.5,1.7)(-1.5,2)( -1.5,-1.7)\qbezier( - 1.5,1.7)(1.7,1.6)(1.7,1.6) \qbezier(-1.6, - 1.7)(1.7,-1.6)(1.7,-1.6)\qbezier( - 1.5,1.7)(1.7,-1.6)(1.7,-1.6)\qbezier(-1.6, -1.7)(1.7,1.6)(1.7,1.6)\qbezier(1.1, -1)(1,-1.5)(1,-1.6)\qbezier(-1,-1.7)( - 0.8,-0.7)(-1.5,-1)\put(-2,1.8){\bf A}\put(1.7,1.7){\bf B} \put(1.7, -1.9){\bf C} \put( - 2, - 1.9){\bf D}\put(0.43,- 1.4){$\bf {40}^{\circ}$}\put( -1.2,-0.7){$\bf {80}^{\circ}$} \end{picture}

\large{\underline{\purple{\frak{ \: Solution:- \: }}}} \\

Let us f1st recall the properties of cyclic quadrilateral. In a cyclic quadrilateral sum of opp. angles is 180° and in a Circle angles made from the same segments are equal.

Here,

\sf \implies \angle ACD = \angle ABD = 40^{\circ} \qquad \bigg \lgroup  Angles\:in\:same\: segment \bigg \rgroup  \\

\sf \implies \angle ABC + \angle ADC = 180^{\circ} \qquad \bigg \lgroup  Opp. \:angles\:of\: cyclic\:quadr. \bigg \rgroup  \\

 \sf where  \footnotesize{\begin{cases}    \\ \sf\angle ADC = 80^{\circ}  \\  \\ \end{cases}}

Substitute this value:-

\sf \implies \angle ABC + \angle ADC = 180^{\circ} \\   \\

\sf \implies \angle ABC + 80^{\circ} = 180^{\circ} \\   \\

\sf \implies \angle ABC= 180^{\circ} - 80^{\circ}  \\   \\

\sf \implies \angle ABC=100^{\circ}  \\   \\

Now,

\sf \dashrightarrow \angle ABC  = \angle ABD + \angle  CBD \\   \\

\sf \dashrightarrow \angle ABC  -  \angle ABD  = \angle  CBD \\   \\

 \sf where  \footnotesize{\begin{cases}\sf\angle ABC  =  100^{\circ}\\  \sf\angle ABD = 40^{\circ}  \end{cases}}

Substituting these values:-

\sf \dashrightarrow \angle CBD =  \angle ABC  -  \angle ABD  \\   \\

\sf \dashrightarrow \angle CBD =  100^{\circ} - 40^{\circ}   \\   \\

\sf \dashrightarrow \angle CBD =  60^{\circ}   \\   \\

\underline{\boxed{\sf \therefore Value \:of\:\angle CBD=60^{\circ}}}

____________________________

\large{\underline{\pink{\frak{ \: Given:- \: }}}} \\

  • \textsf{Ratio of radius of two cylinder is 3:4}\\

\large{\underline{\pink{\frak{ \: Find:- \: }}}} \\

  • \textsf{Ratio of their volume.}\\

\large{\underline{\pink{\frak{ \: Solution:- \: }}}} \\

Let, Radius of 1st cylinder be 'R' = 3

Radius of 2nd Cylinder be 'r' = 4

Height of both cylinder is 'h'

Now, using

 \huge{\underline{\boxed{ \sf Volume \:  of  \: cylinder =  \pi  {r}^{2}h }}}

\underline{\red{\textsf{For 1st cylinder}}}

\implies\sf Volume \:_{1st \: cylinder} =  \pi  {R}^{2}h \\  \\

\implies\sf Volume \:_{1st \: cylinder} =  \pi  {(3)}^{2}h \\  \\

\implies\sf Volume \:_{1st \: cylinder} =  \pi 9h \\  \\

\implies\sf Volume \:_{1st \: cylinder} =  9\pi h \\  \\

\underline{\purple{\textsf{For 2nd cylinder}}}

\implies\sf Volume \:_{2nd \: cylinder} =  \pi  {r}^{2}h \\  \\

\implies\sf Volume \:_{2nd \: cylinder} =  \pi  {(4)}^{2}h \\  \\

\implies\sf Volume \:_{2nd \: cylinder} =  \pi 16h \\  \\

\implies\sf Volume \:_{2nd \: cylinder} =  16\pi h \\  \\

\quad\qquad ________________

\dashrightarrow\sf Ratio_{volumes}= \dfrac{Volume \:_{1st \: cylinder}}{Volume \:_{2nd \: cylinder}} \\  \\

\dashrightarrow\sf Ratio_{volumes}= \dfrac{9 \pi h}{16 \pi h } \\  \\

\dashrightarrow\sf Ratio_{volumes}= \dfrac{9  \not\pi  \not h}{16  \not\pi  \not h } \\  \\

\dashrightarrow\sf Ratio_{volumes}= \dfrac{9}{16} \\  \\

\dashrightarrow\sf Ratio_{volumes}= 9:16 \\  \\

\underline{\boxed{\sf \therefore Ratio\:of\:their\: volumes\:is\:9:16}}

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